1. tính
\(\left(x+2\right)\left(x^2+4\right)\left(x^4+16\right)\)
2. Biến đổi biểu thức sao dưới dạng bình phương tổng or hiệu
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(x^2+2\left(x+1\right)^2+3\left(x-2\right)^2+4\left(x+3\right)^2\)
\(=x^2+2\left(x^2+2x+1\right)+3\left(x^2-4x+4\right)+4\left(x^2+6x+9\right)\)
\(=x^2+2x^2+4x+2+3x^2-12x+12+4x^2+24x+36\)
\(=10x^2+16x+50\)
a) \(\left( {x + 4} \right)\left( {{x^2} - 4x + 16} \right) = {x^3} + {4^3} = {x^3} + 64\)
b) \(\left( {4{x^2} + 2xy + {y^2}} \right)\left( {2x - y} \right) = {\left( {2x} \right)^3} - {y^3} = 8{x^3} - {y^3}\)
\(x^2+2\left(x+1\right)^2+3\left(x+2\right)^2+4\left(x+3\right)^2\)
\(=x^2+2\left(x^2+2x+1\right)+3\left(x^2+4x+4\right)+4\left(x^2+6x+9\right)\)
\(=10x^2+40x+50\)
\(=\left(x^2+10x+25\right)+\left(9x^2+30x+25\right)\)
\(=\left(x+5\right)^2+\left(3x+5\right)^2\)
\(x^2+2\left(x+1\right)^2+3\left(x+2\right)^2+4\left(x+3\right)^2\)
\(=x^2+2\left(x^2+2x+1\right)+3\left(x^2+4x+4\right)+4\left(x^2+6x+9\right)\)
\(=x^2+2x^2+4x+2+3x^2+12x+12+4x^2+24x+36\)
\(=10x^2+40x+50\)
\(=\left(9x^2+30x+25\right)+\left(x^2+10x+25\right)\)
\(=\left(3x+2\right)^2+\left(x+5^2\right)\)
\(x^2+2\left(x+1\right)^2+3\left(x+2\right)^2+4\left(x+3\right)^2\)
\(=x^2+2\left(x^2+2x+1\right)+3\left(x^2+4x+4\right)+4\left(x^2+6x+9\right)\)
\(=x^2+2x^2+4x+2+3x^2+12x+12+4x^2+24x+36\)
\(=10x^2+40x+50\)
Bài 2 :
a ) \(A=\left(a+b+c\right)^2+a^2+b^2+c^2\)
\(A=a^2+b^2+c^2+2ab+2ac+2bc+a^2+b^2+c^2\)
\(A=\left(a^2+2ab+b^2\right)+\left(a^2+2ac+c^2\right)+\left(b^2+2bc+c^2\right)\)
\(A=\left(a+b\right)^2+\left(a+c\right)^2+\left(b+c\right)^2\)
Bài 1:
\(=\left(x^3+4x+2x^2+8\right)\left(x^4+16\right)\)
\(=x^7+16x^3+4x^5+64x+2x^6+32x^2+8x^4+128\)